Given the equation: $ y = -5x^2 + 50x - 129$ Find the parabola's vertex.
Solution: When the equation is rewritten in vertex form like this, the vertex is the point $({h}, {k})$ $ y = A(x - {h})^2 + {k} $ We can rewrite the equation in vertex form by completing the square. First, move the constant term to the left side of the equation: $ \begin{eqnarray} y &=& -5x^2 + 50x - 129 \\ \\ y + 129 &=& -5x^2 + 50x \end{eqnarray} $ Next, we can factor out a $-5$ from the right side: $ y + 129 = -5(x^2 - 10x) $ We can complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $-10$ , so half of it would be $-5$ , and squaring that gives us ${25}$ . Because we're adding the $25$ inside the parentheses on the right where it's being multiplied by $-5$ , we need to add ${-125}$ to the left side to make sure we're adding the same thing to both sides. $ \begin{eqnarray} y + 129 &=& -5(x^2 - 10x) \\ \\ y + 129 + {-125} &=& -5(x^2 - 10x + {25}) \\ \\ y + 4 &=& -5(x^2 - 10x + 25) \end{eqnarray} $ Now we can rewrite the expression in parentheses as a squared term: $ y + 4 = -5(x - 5)^2 $ Move the constant term to the right side of the equation. Now the equation is in vertex form: $ y = -5(x - 5)^2 - 4 $ Now that the equation is written in vertex form, the vertex is the point $({h}, {k})$ $ y = A(x - {h})^2 + {k} $ $ y = -5(x - {(5)})^2 + {(-4)} $ The vertex is $({5}, {-4})$. Be sure to pay attention to the signs when interpreting an equation in vertex form.